Download + R - UCSD Physics

week 7
chapter 32
Fundamentals of Circuits
Resistors: devices that are pieces of conductors
made to have specific resistance
1. a and b
2. a and c
Which of these diagrams
represent the same circuit?
3. b and c
4. a, b, and c
5. a, b, and d
1. a and b
2. a and c
Which of these diagrams
represent the same circuit?
3. b and c
4. a, b, and c
5. a, b, and d
1. ∆V increases by 2V in the direction of I.
What is ∆V across the
2. ∆V decreases by 2V in the direction of I.
unspecified circuit element?
Does the potential increase or 3. ∆V increases by 10V in the direction of I.
decrease when traveling
through this element in the
4. ∆V decreases by 10V in the direction of I.
direction assigned to I?
5. ∆V increases by 26V in the direction of I.
1. ∆V increases by 2V in the direction of I.
What is ∆V across the
2. ∆V decreases by 2V in the direction of I.
unspecified circuit element?
Does the potential increase or 3. ∆V increases by 10V in the direction of I.
decrease when traveling
through this element in the
4. ∆V decreases by 10V in the direction of I.
direction assigned to I?
5. ∆V increases by 26V in the direction of I.
Assume that the voltage of the battery
1) 12 V
is 9 V and that the three resistors are
2) zero
identical. What is the potential
3) 3 V
difference across each resistor?
4) 4 V
5) you need to know the
actual value of R
9V
Assume that the voltage of the battery
1) 12 V
is 9 V and that the three resistors are
2) zero
identical. What is the potential
3) 3 V
difference across each resistor?
4) 4 V
5) you need to know the
actual value of R
Since the resistors are all equal,
the voltage will drop evenly
across the 3 resistors, with 1/3 of
9 V across each one. So we get a
3 V drop across each.
9V
Follow-up: What would be the potential difference if
R= 1 Ω, 2 Ω, 3 Ω
1) 12 V
In the circuit below, what is the
2) zero
voltage across R1?
3) 6 V
4) 8 V
5) 4 V
R 1= 4 Ω
R 2= 2 Ω
12 V
1) 12 V
In the circuit below, what is the
2) zero
voltage across R1?
3) 6 V
4) 8 V
5) 4 V
The voltage drop across R1 has
to be twice as big as the drop
across R2. This means that V1 =
R 1= 4 Ω
R 2= 2 Ω
8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6 Ω) = 2 A, then use
12 V
Ohm’s Law to get voltages.
Follow-up: What happens if the voltage is doubled?
1) 10 A
In the circuit below, what is the
current through R1?
2) zero
3) 5 A
4) 2 A
5) 7 A
R 2= 2 Ω
R 1= 5 Ω
10 V
1) 10 A
In the circuit below, what is the
current through R1?
2) zero
3) 5 A
4) 2 A
5) 7 A
The voltage is the same (10 V) across each
resistor because they are in parallel. Thus,
we can use Ohm’s Law, V1 = I1 R1 to find the
current I1 = 2 A.
R 2= 2 Ω
R 1= 5 Ω
10 V
Follow-up: What is the total current through the battery?
Points P and Q are connected to a
battery of fixed voltage. As more
resistors R are added to the parallel
circuit, what happens to the total
current in the circuit?
1) increases
2) remains the same
3) decreases
4) drops to zero
Points P and Q are connected to a
battery of fixed voltage. As more
resistors R are added to the parallel
circuit, what happens to the total
current in the circuit?
1) increases
2) remains the same
3) decreases
4) drops to zero
As we add parallel resistors, the overall
resistance of the circuit drops. Since V =
IR, and V is held constant by the battery,
when resistance decreases, the current
must increase.
Follow-up: What happens to the current through each resistor?
1. Va = 0V, Vb = −4V, Vc = −10V, Vd = −12V, Ve = −20V
What is the
potential at
points
a to e ?
2. Va = −20V, Vb = −16V, Vc = −10V, Vd = −8V, Ve = 0V
3. Va = −4V, Vb = −6V, Vc = −2V, Vd = −8V, Ve = 0V
4. Va = 4V, Vb = 6V, Vc = 2V, Vd = 8V, Ve = 0V
5. Va = 20V, Vb = 16V, Vc = 10V, Vd = 8V, Ve = 0V
1. Va = 0V, Vb = −4V, Vc = −10V, Vd = −12V, Ve = −20V
What is the
potential at
points
a to e ?
2. Va = −20V, Vb = −16V, Vc = −10V, Vd = −8V, Ve = 0V
3. Va = −4V, Vb = −6V, Vc = −2V, Vd = −8V, Ve = 0V
4. Va = 4V, Vb = 6V, Vc = 2V, Vd = 8V, Ve = 0V
5. Va = 20V, Vb = 16V, Vc = 10V, Vd = 8V, Ve = 0V
1) R1
Which resistor has the
greatest current going
2) both R1 and R2 equally
through it? Assume that all
3) R3 and R4
the resistors are equal.
4) R5
5) all the same
R1
V
R2
R4
R3
R5
1) R1
Which resistor has the
greatest current going
2) both R1 and R2 equally
through it? Assume that all
3) R3 and R4
the resistors are equal.
4) R5
5) all the same
The same current must flow
through left and right
combinations of resistors.
On the LEFT, the current
splits equally, so I1 = I2. On
the RIGHT, more current will
go through R5 than R3 + R4
since the branch containing
R5 has less resistance.
R1
V
R2
R4
R3
R5
Follow-up: Which one has the
smallest voltage drop?
Let’s compute the current in the previous question
R1
V
Ans:
V
I=
Req
R2
Req =
R4
R3
R5
R1 R2
(R3 + R4 )R5
+
R1 + R2
R3 + R4 + R5
Current flows through a
lightbulb. If a wire is now
connected across the bulb,
what happens?
1) all the current continues to flow through
the bulb
2) half the current flows through the wire,
the other half continues through the
bulb
3) all the current flows through the wire
4) none of the above
Current flows through a
lightbulb. If a wire is now
connected across the bulb,
what happens?
1) all the current continues to flow through
the bulb
2) half the current flows through the wire,
the other half continues through the
bulb
3) all the current flows through the wire
4) none of the above
The current divides based on the
ratio of the resistances. If one of the
resistances is zero, then ALL of the
current will flow through that path.
Follow-up: Doesn’t the wire have SOME resistance?
What happens to the voltage across
1) increase
the resistor R1 when the switch is
2) decrease
closed? The voltage will:
3) stay the same
R1
S
V
R3
R2
What happens to the voltage across
1) increase
the resistor R1 when the switch is
2) decrease
closed? The voltage will:
3) stay the same
With the switch closed, the addition of
R1
R2 to R3 decreases the equivalent
S
resistance, so the current from the
battery increases. This will cause an
V
increase in the voltage across R1 .
Follow-up: What happens to the current through R3?
R3
R2
Example: A a problem where Kirchoff’s rules are needed (neither parallel
nor series)
The ammeter in the figure reads 3.0 A. Find I1, I2 and E.
Ans: I1 = 1.0 A, I2 = 2.0 A, E = 15 V
Two lightbulbs operate at 120 V, but one
1) the 25 W bulb
has a power rating of 25 W while the other
2) the 100 W bulb
has a power rating of 100 W. Which one
3) both have the same
has the greater resistance?
4) this has nothing to do
with resistance
Two lightbulbs operate at 120 V, but one
1) the 25 W bulb
has a power rating of 25 W while the other
2) the 100 W bulb
has a power rating of 100 W. Which one
3) both have the same
has the greater resistance?
4) this has nothing to do
with resistance
Since P = V2 / R the bulb with the lower
power rating has to have the higher
resistance.
Follow-up: Which one carries the greater current?
1. Pb > Pa = Pc = Pd
Rank in order, from largest to
smallest, the powers Pa to Pd
dissipated in resistors a to d.
2. Pb = Pc > Pa > Pc
3. Pb = Pd > Pa > Pc
4. Pb > Pc > Pa > Pd
5. Pb > Pd > Pa > Pc
1. Pb > Pa = Pc = Pd
Rank in order, from largest to
smallest, the powers Pa to Pd
dissipated in resistors a to d.
2. Pb = Pc > Pa > Pc
3. Pb = Pd > Pa > Pc
4. Pb > Pc > Pa > Pd
5. Pb > Pd > Pa > Pc
V2/R
(2V)2/R= 4 V2/R
V2/(2R)= (1/2)V2/R
V2/(R/2)= 2 V2/R
1. A = B = C = D
Rank in order, from
brightest to dimmest, the
identical bulbs A to D.
2. A > B > C = D
3. A > C > B > D
4. A > C = D > B
5. C = D > B > A
1. A = B = C = D
Rank in order, from
brightest to dimmest, the
identical bulbs A to D.
2. A > B > C = D
3. A > C > B > D
4. A > C = D > B
5. C = D > B > A
The current through A is the
sum of those through B and
C/D. So A is largest. The
potential across B is the sum
of the potentials across C
and D separately, so the
current through B is larger
than the current through C.
Finally, clearly the same
current goes through C and D.
Two lightbulbs A and B are
connected in series to a
constant voltage source.
When a wire is connected
across B, bulb A will:
1) glow brighter than before
2) glow just the same as before
3) glow dimmer than before
4) go out completely
5) explode
Two lightbulbs A and B are
connected in series to a
constant voltage source.
When a wire is connected
across B, bulb A will:
1) glow brighter than before
2) glow just the same as before
3) glow dimmer than before
4) go out completely
5) explode
Since bulb B is bypassed by the wire,
the total resistance of the circuit
decreases. This means that the current
through bulb A increases.
Follow-up: What happens to bulb B?
The lightbulbs in the circuit below are
1) circuit 1
identical with the same resistance R.
Which circuit produces more light?
(brightness ⇐⇒ power)
2) circuit 2
3) both the same
4) it depends on R
The lightbulbs in the circuit below are
1) circuit 1
identical with the same resistance R.
Which circuit produces more light?
(brightness ⇐⇒ power)
2) circuit 2
In #1, the bulbs are in parallel,
lowering the total resistance of the
circuit. Thus, circuit #1 will draw
a higher current, which leads to
more light, because P = I V.
3) both the same
4) it depends on R
Example: A power line has 0.20 ohms resistance per
kilometer of length. It carries 300 A of current. Find:
(a) the voltage across 1.0 km of the wire, and
(b) the power dissipated in each kilometer of the wire
Ans: (a) 60 V, (b) 18 kW
1. 1 s.
2. 2 s.
The time constant
for the discharge of
this capacitor is
3. 4 s.
4. 5 s.
5. The capacitor doesn’t
discharge because the
resistors cancel each other.
1. 1 s.
2. 2 s.
The time constant
for the discharge of
this capacitor is
3. 4 s.
4. 5 s.
5. The capacitor doesn’t
discharge because the
resistors cancel each other.
Two resistors in series
Equivalent resistance
is R=2Ω+2Ω =4Ω
Time constant
τ = RC = 4 × 1 s = 4 s
In all of the circuits below the
switch is initially open and the
capacitor discharged.
1. b = c = d > a
Rank order the diagrams, form
long to short, by the time it takes
to charge the capacitor once the
switch is closed.
3. b = c > a = d
2. d > b = c > a
4. a = d > b = c
5. b > a = c = d
V
C
S
(a)
R
2R
R
V
C
S
(b)
V
R
2C
S
(c)
2V
C
S
(d)
In all of the circuits below the
switch is initially open and the
capacitor discharged.
1. b = c = d > a
Rank order the diagrams, form
long to short, by the time it takes
to charge the capacitor once the
switch is closed.
3. b = c > a = d
2. d > b = c > a
4. a = d > b = c
5. b > a = c = d
The time constant is RC, independent of V
V
C
S
(a)
R
2R
R
V
C
S
(b)
V
R
2C
S
(c)
2V
C
S
(d)
A battery, a capacitor, and a
resistor are connected in series.
Which of the following affect(s)
the maximum charge stored on
the capacitor?
1. the emf E of the battery
2. the capacitance C of the capacitor
3. the resistance R of the resistor
4. both E and C
5. all three of E, C, and R
A battery, a capacitor, and a
resistor are connected in series.
Which of the following affect(s)
the maximum charge stored on
the capacitor?
1. the emf E of the battery
2. the capacitance C of the capacitor
3. the resistance R of the resistor
4. both E and C
5. all three of E, C, and R
1. the emf E of the battery
A battery, a capacitor, and a resistor
are connected in series. Which of the
following affect(s) the rate at which
the capacitor charges?
2. the capacitance C of the capacitor
3. the resistance R of the resistor
4. both C and R
5. all three of E, C, and R
1. the emf E of the battery
A battery, a capacitor, and a resistor
are connected in series. Which of the
following affect(s) the rate at which
the capacitor charges?
2. the capacitance C of the capacitor
3. the resistance R of the resistor
4. both C and R
5. all three of E, C, and R
The rate at which the capacitor charges is
precisely the current through the resistor,
I=dQ/dt
And
E −t/RC
I= e
R
so the maximum rate is E/R but at later times
the rate also depends on C (from the time constant
RC in the exponent)
Two resistors and an uncharged
capacitor are connected to each
other and then to a battery to form
the circuit shown below.
Long after this is done, the
currents I1 and I2 through the
resistors R1 and R2, respectively,
and the potential V across the
capacitor are
1. I1 ≠ 0, I2 = 0, V = 0
2. I1 = 0, I2 = 0, V = 0
3. I1 ≠ 0, I2 = 0, V ≠ 0
4. I1 = 0, I2 ≠ 0, V ≠ 0
5. I1 ≠ 0, I2 ≠ 0, V ≠ 0
R1
+
–
R2
C
Two resistors and an uncharged
capacitor are connected to each
other and then to a battery to form
the circuit shown below.
Long after this is done, the
currents I1 and I2 through the
resistors R1 and R2, respectively,
and the potential V across the
capacitor are
1. I1 ≠ 0, I2 = 0, V = 0
2. I1 = 0, I2 = 0, V = 0
3. I1 ≠ 0, I2 = 0, V ≠ 0
4. I1 = 0, I2 ≠ 0, V ≠ 0
5. I1 ≠ 0, I2 ≠ 0, V ≠ 0
Once the capacitor is charged
no more current flows through
it. The two resistors form a
series circuit with the battery,
so the current through them is
non-zero. Also, the capacitor is
charged, so V ≠ 0
R1
+
–
R2
C
Same situation: Two resistors and an
uncharged capacitor are connected
to each other and then to a battery to
form the circuit shown below.
Long after this is done, the currents
I1 and I2 through the resistors R1 and
R2, respectively, and the potential V
across the capacitor satisfy
1. I2 = V / R2
2. I1 = I2
3. I1 =E / (R1+R2)
4. all of the above
5. no more than two of the above
R1
+
–
R2
C
Same situation: Two resistors and an
uncharged capacitor are connected
to each other and then to a battery to
form the circuit shown below.
Long after this is done, the currents
I1 and I2 through the resistors R1 and
R2, respectively, and the potential V
across the capacitor satisfy
1. I2 = V / R2
2. I1 = I2
3. I1 =E / (R1+R2)
4. all of the above
5. no more than two of the above
R1
+
–
R2
C
Example: (a) What is the charge in the capacitor a long time after this
circuit has been connected?
(b) What is the current through the battery when this is initially set up,
if the capacitor is initially discharged?
(“initially set up” means... imagine having a switch connected to the
battery, with the switch initially open)
C
R
+
–
Ans: (a) C E/2, (b) E/R
R
more slides
1) increases
What happens to the voltage
across the resistor R4 when the
2) decreases
3) stays the same
switch is closed?
R1
S
V
R3
R2
R4
1) increases
What happens to the voltage
across the resistor R4 when the
2) decreases
3) stays the same
switch is closed?
We just saw that closing the switch
causes an increase in the voltage
across R1 (which is VAB). The
voltage of the battery is constant,
so if VAB increases, then VBC must
decrease!
A
R1
B
S
V
R3
R2
C
Follow-up: What happens to the current through R4?
R4
Two space heaters in your living
room are operated at 120 V. Heater 1
has twice the resistance of heater 2.
Which one will give off more heat?
1) heater 1
2) heater 2
3) both equally
Two space heaters in your living
room are operated at 120 V. Heater 1
has twice the resistance of heater 2.
Which one will give off more heat?
1) heater 1
2) heater 2
3) both equally
Using P = V2 / R, the heater with the smaller resistance
will have the larger power output. Thus, heater 2 will
give off more heat.
Follow-up: Which one carries the greater current?
The three lightbulbs in the circuit all have the
same resistance of 1 Ω . By how much is the
brightness of bulb B greater or smaller than
the brightness of bulb A? (brightness ⇐⇒
power)
1) twice as much
2) the same
3) 1/2 as much
4) 1/4 as much
5) 4 times as much
10 V
The three light bulbs in the circuit all have the
same resistance of 1 Ω . By how much is the
brightness of bulb B greater or smaller than
the brightness of bulb A? (brightness ⇐⇒
power)
1) twice as much
2) the same
3) 1/2 as much
4) 1/4 as much
5) 4 times as much
We can use P = V2/R to compare the power:
PA = (VA)2/RA = (10 V) 2/1 Ω = 100 W
PB = (VB)2/RB = (5 V) 2/1 Ω = 25 W
Follow-up: What is the total current in the circuit?
10 V
1. bulb B glows more brightly
In the circuit shown, the two
bulbs A and B are identical.
Compared to bulb A,
2. bulb B glows less brightly
3. bulb B glows just as brightly
4. answer depends on whether the mobile
charges in the wires are positively or
negatively charged
1. bulb B glows more brightly
In the circuit shown, the two
bulbs A and B are identical.
Compared to bulb A,
2. bulb B glows less brightly
3. bulb B glows just as brightly
4. answer depends on whether the mobile
charges in the wires are positively or
negatively charged
Bulbs are in series, hence same
current I through them.
Brightness is related to power
dissipated P = I2R; bulbs are
identical, so R is the same, and
since I is also the same, they
dissipate same power, hence same
brightness.
In the circuit shown in (a), the two bulbs A
and B are identical. Bulb B is removed and
the circuit is completed as shown in (b).
Compared to the brightness of bulb A in (a),
bulb A in (b) is
1. brighter
2. less bright
3. just as bright
4. any of the above, depending
on the rated wattage of the bulb
In the circuit shown in (a), the two bulbs A
and B are identical. Bulb B is removed and
the circuit is completed as shown in (b).
Compared to the brightness of bulb A in (a),
bulb A in (b) is
1. brighter
2. less bright
3. just as bright
4. any of the above, depending
on the rated wattage of the bulb